Based on this, where does the projection map send an arbitrary open set?

For (4), use (3): take the product of the Sa and show that it's closure is X.

Thank you very much for your reply. Here are my answers so far based on your suggestions.

1. The product topology [tex]\tau[/tex] is generated from base [tex]\mathfrak{B}[/tex] consisting of product sets [tex]\displaystyle \prod_{a \in A} U_a[/tex] where only finitely many factors are not [tex]X_a[/tex] and the remaining factors are open sets in [tex]X_a[/tex]. Therefore the project [tex]p_a[/tex] projects an open set [tex]S \subseteq X[/tex] to either [tex]X_a[/tex] or some open subset [tex]S_a \subset X_a[/tex].

***This could be wrong, if [tex]|A| = |\mathbb{N}|[/tex], then the cartesian product does not have to be countable. So what is the set separable? Should the question say If [tex]|A| < |\mathbb{N}|[/tex] and each [tex]|X_a|[/tex] is separable, prove that [tex]|X|[/tex] is separable?